I am trying to link an image using the same name as a $_GET variable, example bellow:
The $_GET
$venue = $_GET['venue'];
Is it possible to use '$venue' as the image src example being something like
<div id="imgbox">
<img src="$venue.jpg" alt="venueimage" height="150" width="250">
</div>
My attempts so far have been unsuccessful, is it possible in a similar way to this or is there an alterantive?
Thankyou
You forgot your PHP tags and echo statement:
<div id="imgbox">
<img src="<?php echo $venue; ?>.jpg" alt="venueimage" height="150" width="250">
</div>
or shorthand:
<div id="imgbox">
<img src="<?= $venue; ?>.jpg" alt="venueimage" height="150" width="250">
</div>
As pointed out by Quentin we should take this a step further and sanitize our output:
<div id="imgbox">
<img src="<?= htmlspecialchars($venue); ?>.jpg" alt="venueimage" height="150" width="250">
</div>
You should be able to use <img src="<?php echo $venue; ?>.jpg" alt="venueimage" height="150" width="250">
I forgot to sanitize the code as well. You would need to add:
<img src="<?php echo htmlspecialchars($venue); ?>.jpg" alt="venuimage" height="150" width="250">
Related
After retrieve image path from database I want to pass it to <img> tag to show up,
I read this solution, but I do not want to echo it, I want to assign it as src for image tag
I tried this:
$image = mysqli_fetch_array($user_images);
$img1 = $image['file_path'];
$imageData = base64_encode(file_get_contents($img1));
$src = 'data: '.mime_content_type($img1).';base64,'.$imageData;
How can I assign $src to image, I tried this but no luck :(
<img id="img1" width="150" height="150" src="' . $src . '"/>
Thanks in advance.
Inside the HTML you still need to echo the value of $src. This should work:
<img id="img1" width="150" height="150" src="<?php echo $src; ?>"/>
Try with PHP tags like below:-
<img id="img1" width="150" height="150" src="<?php echo $src;?>"/>
Reasonable solution I can think of is
<img id="img1" width="150" height="150" src="<?php echo $src; ?>"/>
but the code where you get the image source has to be above the image tag, so you have the $src.
You are using PHP language and in Php, you can't pass variable directly in the Tag you have to set a variable in the PHP environment.
To create Php environment you can right <?php ?> or you can use to echo variable <?= ?> between these PHP tags you can pas your variables to echo and assign or anything else what you want to do.
<img id="img1" width="150" height="150" src="<?= $src ?>"/>
$images['result'][0]['image_path'] - here I have path of image. It is in public folder.
<img width="150" height="150" alt="150x150" src="<?php BASEPATH."../public/" echo $images['result'][0]['image_path'];?>" />
Can anyone help me to display this image?
Try This,
<img width="150" height="150" alt="150x150" src="<?php echo base_url('your/Folder/Structure/').$images['result'][0]['image_path'];?>" />
First concate your file name with path which causing the issue.
base_url() will also help you to fetch/show your image
I've encountered this issue a couple times but have always found a "hack" way around it. Is there a special way to link an image in a WordPress template to an outside url beyond the typical a href tag? Here's the images I'm trying to link to outside urls:
<div class="socialMedia">
Follow Us: <br />
<img src="<?php echo get_bloginfo('template_url') ?>/images/facebook.png" alt="facebook"/>
<img src="<?php echo get_bloginfo('template_url') ?>/images/twitter.png" alt="twitter"/>
<img src="<?php echo get_bloginfo('template_url') ?>/images/googleplus.png" alt="google plus"/>
<img src="<?php echo get_bloginfo('template_url') ?>/images/instagram.png" alt="instagram"/>
</div><!--.socialMedia-->
<div class="socialMedia">
Follow Us: <br />
<a href="<?php echo get_bloginfo('template_url') ?>/images/facebook.png" target="_blank" > <img src="<?php echo get_bloginfo('template_url') ?>/images/facebook.png" alt="facebook"/> </a>
</div><!--.socialMedia-->
did you mean it!!?
<li><a href='#'>Core Transformation</a>
<img alt='arrow' src=**'http://localhost/wordpress1/wp-content/themes/twentytwelve/images/header-triangle.png'** /></li>
<li><a href=**'#'**>LINKS</a></li>
<li><a href=**'#'**>Contact Us</a></li>
How can I replace the src and href using the site_url() function?
You can use the following ways :
<img src="<?php bloginfo('template_url'); ?>/images/image.jpg" />
<img src="<?php echo get_template_directory_uri();?>/images/image.jpg" />
<img src="<?php home_url();?>/images/image.jpg" />
<img src="<?php echo site_url();?>/images/image.jpg"/>
you can do it same for href
More details refer codex
<img src="<?php echo site_url();?>/images/image.jpg"/>
or a better option would be:
<img src="<?php bloginfo('template_url'); ?>/images/image.jpg" />
You can do similarly for href
site_url() is used to append text on url.. so if you want to navigate to your directory and fetch something you can use it like this..
site_url('/images/default.jpg');
you can replace your src with following way:
<img src="<?php bloginfo('template_url'); ?>/images/yourimagename.extension" />
for more information about wordpress go to http://www.wpbeginner.com/wp-themes/wordpress-theme-cheat-sheet-for-beginners
that will help you more in future.
Still you getting any issues with that,then comment it.
Not even sure if methods is the correct terminology...
Here is the original working code:
<a href="<?php bloginfo('url'); ?>">
<img src="<?php bloginfo('stylesheet_directory'); ?>/images/logo.png" alt="Polished Logo" id="logo"/></a>
<img src="<?php bloginfo('stylesheet_directory'); ?>/images/separator.png" width="2" height="59" alt="Line" class="logo_line"/>
<p id="logo_title"><?php bloginfo('description'); ?></p>
I wanted it to only execute on the homepage, so I wrote this:
<?
if ( $_SERVER["REQUEST_URI"] == '/' ){
echo '<a href="'.bloginfo('url').'">
<img src="'.bloginfo('stylesheet_directory').'/images/logo.png" alt="Polished Logo" id="logo"/></a>
<img src="'.bloginfo('stylesheet_directory').'/images/separator.png" width="2" height="59" alt="Line" class="logo_line"/>
<p id="logo_title">'.bloginfo('description').'</p>';
}
?>
But it outputs the bloginfo() and the other declarations completely outside the html tags I have created. For instance, with bloginfo('stylesheet_directory') it will display the directory outside the IMG tags I created.
Any ideas? Apparently my syntax isn't correct or something....
bloginfo function directly echoes the output. In this case you should use get_bloginfo to add the returned value to the string and echo the complete string. I believe this will work
<?php
if ( $_SERVER["REQUEST_URI"] == '/' ) {
echo '<a href="'.get_bloginfo('url').'">
<img src="'.get_bloginfo('stylesheet_directory').'/images/logo.png" alt="Polished Logo" id="logo"/></a>
<img src="'.get_bloginfo('stylesheet_directory').'/images/separator.png" width="2" height="59" alt="Line" class="logo_line"/>
<p id="logo_title">'.get_bloginfo('description').'</p>';
}
?>
Here is a better alternative:
<?php if ( $_SERVER["REQUEST_URI"] == '/' ) { ?>
<a href="<?php bloginfo('url') ?>">
<img src="<?php bloginfo('stylesheet_directory') ?>/images/logo.png" alt="Polished Logo" id="logo"/>
</a>
<img src="<?php bloginfo('stylesheet_directory') ?>/images/separator.png" width="2" height="59" alt="Line" class="logo_line"/>
<p id="logo_title"><?php bloginfo('description') ?></p>
<?php } ?>
I also suggest using the is_home() function provided by wordpress to check for the homepage instead of checking the $_SERVER['REQUEST_URI'] value.
bloginfo() outputs data with echo and returns nothing, so instead of trying to concatenate everything, just output in sequence, e.g.
echo '<a href="';
bloginfo('url');
echo '"><img src="';
bloginfo('stylesheet_directory');
//etc...
Ugly I know, but see answer by Nithesh for a possible alternative.
if you want to get the template path without auto echoing it by the bloginfo() function use:
get_bloginfo( 'stylesheet_directory', 'display' )